Let $\lambda=\min(f/V)$. Then $u-\lambda$ satisifies $$-\Delta(u-\lambda)+V(u-\lambda)=f-\lambda V\ge 0.$$ Hence $u\ge\lambda$ by the maximum principle.

Here are a couple of ideas:

1 This yields a lower bound, but it is not in terms of a norm of f:

Let $\lambda=\min(f/V)$. Then $u-\lambda$ satisifies $$-\Delta(u-\lambda)+V(u-\lambda)=f-\lambda V\ge 0.$$ Hence $u\ge\lambda$ by the maximum principle.

2 Assume V is bounded, and let $V_M$ be its maximum: Let $v$ be the solution of $-\Delta v+V_Mv=f$. Then $$-\Delta(u-v)+V(u-v)=(V_M-V)v\ge 0,$$ so $u\ge v$. If the geometry of the domain is simple, you may be able to determine the Green's function for $-\Delta+V_M$ explicitly.